# Some help required

• March 12th, 2009, 08:27 AM
Manynames
Some help required
I need to know if the following has been deducted correctly. I am renown for making mathematical mistakes :twisted: but here it goes:

Prove that inertia can be given as the producct of inertial mass plus its relativistic mass. The case proven:

the small signature for the momentum is used as a trace of information, thus:

by rearranging:

So if you added the inertial mass in question (given any constant value) with the addition of some relativistic mass, then we have:

So naturally, we have used the upper case for momentum a relativistic gamma relation. If we remove the zero value of the gamma relations on both sides, we are left with this simple formula:

so we have an inertial mass from inviting a relativistic velocity minus its equivalent inertia. From here, the equation can be interpreted as how something has an inertial mass, and that is when you take the same quantity and take it from the relativistic velocity, which zero, so all of this simplifies to a proof of Einsteins Weak Equivalence, so that:

Inertial mass is the same as the quantity of mass as well. Now since we are talking about mass and relativistic mass relationships, i now want to derive for density relationships. So following these derivatives, we use wavelength of a thing as important as upon the sqaure of any wave, a density indeed can be given, but for the sake of simplicity, let us not get too entwined into that. Let's assume though the following, albeit as simple as it is:

solving yet again for the left hand side (warning, do not get mixed up where v is a volume.

which solves to

This should be seen as simple equating and rearranging, knowing that . Does this all seem reasonable,

Manynames
• March 12th, 2009, 08:50 AM
Manynames
Re: Some help required
Quote:

Originally Posted by Manynames
I need to know if the following has been deducted correctly. I am renown for making mathematical mistakes :twisted: but here it goes:

Prove that inertia can be given as the producct of inertial mass plus its relativistic mass. The case proven:

the small signature for the momentum is used as a trace of information, thus:

by rearranging:

So if you added the inertial mass in question (given any constant value) with the addition of some relativistic mass, then we have:

So naturally, we have used the upper case for momentum a relativistic gamma relation. If we remove the zero value of the gamma relations on both sides, we are left with this simple formula:

so we have an inertial mass from inviting a relativistic velocity minus its equivalent inertia. From here, the equation can be interpreted as how something has an inertial mass, and that is when you take the same quantity and take it from the relativistic velocity, which zero, so all of this simplifies to a proof of Einsteins Weak Equivalence, so that:

Inertial mass is the same as the quantity of mass as well. Now since we are talking about mass and relativistic mass relationships, i now want to derive for density relationships. So following these derivatives, we use wavelength of a thing as important as upon the sqaure of any wave, a density indeed can be given, but for the sake of simplicity, let us not get too entwined into that. Let's assume though the following, albeit as simple as it is:

solving yet again for the left hand side (warning, do not get mixed up where v is a volume.

which solves to

This should be seen as simple equating and rearranging, knowing that . Does this all seem reasonable,

Manynames

There is actually more i needed to be looked over, cheers :D

Since any wavelength of a particle photon moves along a null path, even the absence of a fourth time vector can surely be true under:

which clears up under:

If there is a reason to think of a ''critical energy'' for the density of a photon, i am now to assume that this would be best thought of from a particle moving in a field. This would be best described under

so we are thus referring to the gravitational charge of a moving particle given by the presence of a gravitational field charge. So one can think of this as being the same as the ''gravitational charge of a particle,'' i will denote for pure simplicity for summing equations as so that to the nth state, we can evaluate some kind of gravitational density due to the charge.

so we can now use a summation method:

From here, i hope it might be obvious that the state is the ''critical density due to the gravitational charge. This would suspect me to believe that the gravitational charge itself must in some way be the quantity of inertia, without the impression of a relativistic term.

• March 12th, 2009, 01:11 PM
Manynames
Can anyone here help? I'd appreciate any, thanks. Just looking for mistakes, and i can't find any, but i'm notorious for not spotting them, so any help would be good

:wink: